Basis, Spanning set, Linear combination I am confused with the linear combination concept, see question below.

Consider the three vectors $u = (5, -4, 7)$, $v=(2,-1,3)$ and $w=(-1,2-1)$ in $\mathbb{R}^3$.
Answer $\textbf{True or False}$ to each of the following statements AND briefly explain why it is True or False. 
a) $u$ is a linear combination of $v$ and $w$ 
b) $\{u,v,w\}$ are linearly independent. 
c)  $\{u,v,w\}$ form a basis for $\mathbb{R}^3$ .
d) $span\{u,v,w\}$ forms a line in $\mathbb{R}^3$ . 
e) $dim_\mathbb{R}\;( span\{u,v,w\} )=3$
f) The following matrix is not invertible: 
$$\begin{bmatrix}     5 &  2& -1\\ 
       -4 &-1& 2\\ 
        7  &3& -1\end{bmatrix}$$

My Try:
a)  I realised there were 3 equations:
$\quad2C_1 – C_2 = 5$
$\quad-C_1 + 2C_2 = -4$
$\quad3C_1 – C_2 = 7$
And when I used Gauss-Jordon, the matrix became inconsistent, so the answer is no,
b)  False as because, there was no solution and the equations did not solve to have the trivial solution
c)  I am not really sure what this means, in order for them to be a basis of $\mathbb{R}^3$ does it mean they have to be linearly independent and because they aren’t. This statement is false?
d)  The spanning set of the three vectors forms a line in $\mathbb{R}^3$, well they are not a linear combination of each other so I think this is false?
e)    I have no idea,
f)  True because if there is no solution the determinant = 0 and that means it is not invertible
Are my answers correct, and what is the answer to part e, please provide explanation.
 A: For $a$, you should have gotten the solution $c_1=2, c_2=-1$.
You're correct for $b$ and $c$. For $d$, note that the spanning set is going to have dimension $0$, $1$, $2$, or $3$. We know it won't be $0$, and it can't be $3$ since $\{u,v,w\}$ isn't a basis for $\mathbb R^3$. Further, the spanning set doesn't have dimension $1$, since the vectors clearly do not fit on the same line (since, for example, $v$ is not a scalar multiple of $v$). Therefore, the span of these vectors forms a plane. 
$e$ is false; if the dimension of the spanning set was $3$, then the spanning set would just be $\mathbb R^3$. Since there are only three vectors, this would mean that $\{u,v,w\}$ is a basis for $\mathbb{R}^3$, which it isn't.
You're good for $f$. 
A: I will not write about this specific vectors, rather something I think is more important, and is that some of the questions are 'equal'-some of them are "iff" statements (actually all except 'd', since they can be linearly dependent but form a 2-dimensional field).
If you have 3 vectors for example, and they are all linearly independent, than the dimension of the span is 3, and they span $\mathbb R^3$, same goes for any number of vectors and dimension and $\mathbb R^n$ respectively.
This will help you with question 'e'.
If you need further help, write here in a comment and I'll explain.
